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A circle is passing through the points A (1, 1) and B (1, 3) and the bisector of first and third quadrant is normal to it, then its area, in square units, isA. `2pi`B. `4pi`C. `(pi)/(2)`D. none of these |
Answer» Correct Answer - A It is given that y=x is a normal to the circle. So, its centre lies on y=x. C (t, t) be the coordinates of the centre. AB is a chord of the circle. So, the perpendicular bisector of AB passes through the centre of the circle. The equation of the perpendicular bisector of AB is y = 2. Thus, the coordinates of the centre are (2, 2). `:.` Radius = CA=`sqrt((2-1)^(2)+(2-1)^(2))=sqrt(2)`. `:.` Area of the circle `= 2 pi` sq. units. |
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